Question: Let $y=\cot(x)$. What is the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{2}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $-1$ (Choice C) C $1$ (Choice D) D $\dfrac12$
Explanation: Let's first find $\dfrac{dy}{dx}$. Then, we can evaluate it at $x=\dfrac{\pi}{2}$. Recall that the derivative of $\cot(x)$ is $-\dfrac{1}{\sin^2(x)}$, or $-\csc^2(x)$. Put another way, $\dfrac{d}{dx}[\cot(x)]=-\dfrac{1}{\sin^2(x)}=-\csc^2(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{\pi}{2}}$ : $\begin{aligned} &\phantom{=}-\dfrac{1}{\sin^2\left({\dfrac{\pi}{2}}\right)} \\\\ &=-\dfrac{1}{\left(1\right)^2} \\\\ &=-1 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{2}$ is $-1$.